Re: Issues modifying a local variable inside of a subfunction

[Hunter Wittenborn]

> That's right. The 'declare -g' command operates on the global

> scope, not the immediately enclosing scope (unless that *is* the

> global scope). 



> However, since you're using 'declare' to do indirect assignments,

> you can modify local variables and global variables but nothing in

> between. 



Alright, that makes sense. That was where I was getting tripped up, I just 
needed confirmation on it.



> test1() {

> printf -v "$1" '%s' 1

> }



Completely forgot printf can do this! I haven't used it much in that manner 
yet, so I'm assuming that's I didn't think of it. Regarding solutions though, 
that method works out beautifully for me.



Thanks! That looks like it's going to solve my issue just fine.



Thanks for all the inspiring words as well, they're appreciated! :)



---

Hunter Wittenborn

mailto:hunter@hunterwittenborn.com

https://www.hunterwittenborn.com

https://github.com/hwittenborn






---- On Mon, 11 Oct 2021 22:18:43 -0500 Lawrence Velázquez <vq@larryv.me> wrote 
----



On Mon, Oct 11, 2021, at 10:08 PM, Hunter Wittenborn wrote: 
> I'm currently having some issues modifying shell variable that have 
> been declared with 'local' via subfunctions in a given function. 
> 
> The following gives what my actual script is trying to do: 
> 
> """ 
> #!/usr/bin/env bash 
> test1() { 
>   variable="${1}" 
>   declare -g "${variable}=1" 
> } 
> 
> test2() { 
>   local number=0 
>   test1 number 
>   echo "${number}" 
> } 
> 
> test2 
> """ 
> 
> Why is this outputting "0" instead of "1"? After reading over the Bash 
> man pages I was thinking 'declare -g' might be assigning to a variable 
> reference outside of that made by my 'local' command 
 
That's right.  The 'declare -g' command operates on the global 
scope, not the immediately enclosing scope (unless that *is* the 
global scope). 
 
 % cat /tmp/foo1.bash; echo 
 test1() { 
 declare -g "$1"=1 
 } 
 
 test2() { 
 local number=0 
 test1 number 
 printf 'test2:  %s\n' "$number" 
 } 
 
 number=2 
 test2 
 printf 'global: %s\n' "$number" 
 
 % bash /tmp/foo1.bash 
 test2:  0 
 global: 1 
 
> but at that point I can't figure out how I can modify the value 
> of 'number' while inside a subfunction of the 'test2()' function. 
 
You're conflating a couple of things here.  You can easily modify 
test2's 'number' from within 'test1'. 
 
 % cat /tmp/foo2.bash; echo 
 test1() { 
 number=1 
 } 
 
 test2() { 
 local number=0 
 test1 
 printf 'test2:  %s\n' "$number" 
 } 
 
 number=2 
 test2 
 printf 'global: %s\n' "$number" 
 
 % bash /tmp/foo2.bash 
 test2:  1 
 global: 2 
 
However, since you're using 'declare' to do indirect assignments, 
you can modify local variables and global variables but nothing in 
between. 
 
> Is this a bug, or is it just behaving as expected from how I described 
> above? If the latter, is there any way to accomplish what I'm wanting 
> to do? 
 
Ideally you would avoid doing indirect assignments.  Without context, 
it's hard to make a holistic suggestion.  You can avoid the scoping 
issue by using a technique that doesn't involve 'declare'.  For 
example: 
 
 % cat /tmp/foo3.bash; echo 
 test1() { 
 printf -v "$1" '%s' 1 
 } 
 
 test2() { 
 local number=0 
 test1 number 
 printf 'test2:  %s\n' "$number" 
 } 
 
 number=2 
 test2 
 printf 'global: %s\n' "$number" 
 
 % bash /tmp/foo3.bash 
 test2:  1 
 global: 2 
 
There are many such techniques, and each one comes with its own 
baggage.  Have fun! 
 
https://mywiki.wooledge.org/BashFAQ/006#Assigning_indirect.2Freference_variables
 
 
-- 
vq